matrices problem Watch

bigmansouf
Badges: 20
Rep:
?
#1
Report Thread starter 3 weeks ago
#1
Question:

Under a certain transformation the image of the point (x,y) is (X,Y) where \begin{pmatrix} X\\  Y \end{pmatrix} = \begin{pmatrix} 1 & 4\\  2 & 3 \end{pmatrix} \begin{pmatrix} x\\  y \end{pmatrix} . This transformation maps any point on the line  y = mx onto another point on the line  y=mx. Find the two possible values of m.


My attempt:
I tried to look at the the effect of the transformation on the unit square  \begin{pmatrix} X\\  Y \end{pmatrix} = \begin{pmatrix} 1 & 4\\  2 & 3 \end{pmatrix} \begin{pmatrix} 0 &1  &1  &0 \\  0 &0  & 1 & 1 \end{pmatrix}
 = \begin{pmatrix} 0 &1  &5  &4 \\  0 &2  & 5 & 3 \end{pmatrix}

Since it says y= mx
i thought it had something to do with reflection in the line y=mx where m = tan \alpha
I have tried to calculate the angle  \alpha between i.e ( 1,0) and its image (1,2) or (0,1) (4,3) but it could not figure it out.

I would like to know if my approach is wrong or right?
Please can i have help on finding the two values of m
Last edited by bigmansouf; 3 weeks ago
0
reply
RDKGames
  • Study Forum Helper
Badges: 20
Rep:
?
#2
Report 3 weeks ago
#2
(Original post by bigmansouf)
Question:

Under a certain transformation the image of the point (x,y) is (X,Y) where \begin{pmatrix} X\\  Y \end{pmatrix} = \begin{pmatrix} 1 & 4\\  2 & 3 \end{pmatrix} \begin{pmatrix} x\\  y \end{pmatrix} . The transformation maps any point on the line  y = mx . Find the two possible values of m.


My attempt:
I tried to look at the the effect of the transformation on the unit square  \begin{pmatrix} X\\  Y \end{pmatrix} = \begin{pmatrix} 1 & 4\\  2 & 3 \end{pmatrix} \begin{pmatrix} 0 &1  &1  &0 \\  0 &0  & 1 & 1 \end{pmatrix}
 = \begin{pmatrix} 0 &1  &5  &4 \\  0 &2  & 5 & 3 \end{pmatrix}

Since it says y= mx
i thought it had something to do with reflection in the line y=mx where m = tan \alpha
I have tried to calculate the angle  \alpha between i.e ( 1,0) and its image (1,2) or (0,1) (4,3) but it could not figure it out.

I would like to know if my approach is wrong or right?
Please can i have help on finding the two values of m
Just evaluate

\begin{pmatrix} 1 & 4 \\ 2 & 3 \end{pmatrix} \begin{pmatrix} x \\ mx \end{pmatrix}


Though the question doesn't really make sense. It doesn't give any conditions to work with. Is it meant to say that every point on this line maps to itself?
1
reply
bigmansouf
Badges: 20
Rep:
?
#3
Report Thread starter 3 weeks ago
#3
(Original post by RDKGames)
Just evaluate

\begin{pmatrix} 1 & 4 \\ 2 & 3 \end{pmatrix} \begin{pmatrix} x \\ mx \end{pmatrix}


Though the question doesn't really make sense. It doesn't give any conditions to work with. Is it meant to say that every point on this line maps to itself?
I edited it to include the part i missed

'This transformation maps any point on the line y = mx onto another point on the line y=mx'
0
reply
RDKGames
  • Study Forum Helper
Badges: 20
Rep:
?
#4
Report 3 weeks ago
#4
(Original post by bigmansouf)
I edited it to include the part i missed

'This transformation maps any point on the line y = mx onto another point on the line y=mx'
Cool, so just use the fact that (Y,X) lies on y=mx... so Y=mX must hold.
0
reply
Ryanzmw
Badges: 16
Rep:
?
#5
Report 3 weeks ago
#5
(Original post by bigmansouf)
Question:

Under a certain transformation the image of the point (x,y) is (X,Y) where \begin{pmatrix} X\\  Y \end{pmatrix} = \begin{pmatrix} 1 & 4\\  2 & 3 \end{pmatrix} \begin{pmatrix} x\\  y \end{pmatrix} . This transformation maps any point on the line  y = mx onto another point on the line  y=mx. Find the two possible values of m.


My attempt:
I tried to look at the the effect of the transformation on the unit square  \begin{pmatrix} X\\  Y \end{pmatrix} = \begin{pmatrix} 1 & 4\\  2 & 3 \end{pmatrix} \begin{pmatrix} 0 &1  &1  &0 \\  0 &0  & 1 & 1 \end{pmatrix}
 = \begin{pmatrix} 0 &1  &5  &4 \\  0 &2  & 5 & 3 \end{pmatrix}

Since it says y= mx
i thought it had something to do with reflection in the line y=mx where m = tan \alpha
I have tried to calculate the angle  \alpha between i.e ( 1,0) and its image (1,2) or (0,1) (4,3) but it could not figure it out.

I would like to know if my approach is wrong or right?
Please can i have help on finding the two values of m
Isn't this equivalent to finding the eigenvectors of the matrix?

Because then if v is an eigenvector of M, Mv = kv, i.e. the transformation under M is colinear to the vector you started with, which is essentially what this mapping line to line business is saying. If you're not sure how to calculate eigenvectors of a matrix, there is a wealth of information line describing this process.

Spoiler:
Show
The eigenvectors are (1,1)^T, (-2,1)^T so the lines would be y = x and y = -1/2 * x (I think)
Last edited by Ryanzmw; 3 weeks ago
0
reply
bigmansouf
Badges: 20
Rep:
?
#6
Report Thread starter 3 weeks ago
#6
 \begin{pmatrix}

X\\ 

Y

\end{pmatrix} = \begin{pmatrix}

1 &4 \\ 

2& 3

\end{pmatrix} \begin{pmatrix}

X\\ 

Y

\end{pmatrix}

I learnt that the basics of egienvectors is  A\vec{v}=\lambda \vec{v}
 A\vec{v} - \lambda I \vec{v} = 0
 (A - \lambda I) \vec{v} = 0
 det(A - \lambda I) \vec{v}=0

 \lambda \begin{pmatrix}

x\\ 

y\end{pmatrix} = \begin{pmatrix}

1 &4 \\ 

2& 3

\end{pmatrix}\begin{pmatrix}

x\\ 

y

\end{pmatrix}

 \left (\left \begin{pmatrix} 1 &4 \\  2& 3 \end{pmatrix}-\begin{pmatrix} \lambda & 0 \\  0 & \lambda \end{pmatrix}  \right )\begin{pmatrix} x\\  y \end{pmatrix}=0

 det(A - \lambda I) \vec{v}=0
therefore
  ((1-\lambda )(3-\lambda )-8)=0
  \lambda =5  or \lambda = -1

I sub each  \lambda into the matrix;
  \begin{pmatrix} 1-\lambda & 4\\  2 & 3- \lambda \end{pmatrix}\begin{pmatrix} x\\  y \end{pmatrix} = 0

when  \lambda =5
 \begin{pmatrix} -4 & 4\\  2 & -2 \end{pmatrix}\begin{pmatrix} x\\  y \end{pmatrix} = 0

 -4x+4y=0
 2x-2y=0
thus y=x

when  \lambda = -1
 \begin{pmatrix} 2 & 4\\  2 & 4 \end{pmatrix}\begin{pmatrix} x\\  y \end{pmatrix} = 0
 2x +4y =0
  2x+ 4y = 0
thus   y = -\frac{1}{2}x


Did i follow the right approach?


(Original post by Ryanzmw)
Isn't this equivalent to finding the eigenvectors of the matrix?

Because then if v is an eigenvector of M, Mv = kv, i.e. the transformation under M is colinear to the vector you started with, which is essentially what this mapping line to line business is saying. If you're not sure how to calculate eigenvectors of a matrix, there is a wealth of information line describing this process.

Spoiler:
Show


The eigenvectors are (1,1)^T, (-2,1)^T so the lines would be y = x and y = -1/2 * x (I think)
Last edited by bigmansouf; 3 weeks ago
1
reply
Ryanzmw
Badges: 16
Rep:
?
#7
Report 3 weeks ago
#7
Yep. Do you see how this relates to the question?
(Original post by bigmansouf)
 \begin{pmatrix}

X\\ 

Y

\end{pmatrix} = \begin{pmatrix}

1 &4 \\ 

2& 3

\end{pmatrix} \begin{pmatrix}

X\\ 

Y

\end{pmatrix}

I learnt that the basics of egienvectors is  A\vec{v}=\lambda \vec{v}
 A\vec{v} - \lambda I \vec{v} = 0
 (A - \lambda I) \vec{v} = 0
 det(A - \lambda I) \vec{v}=0

 \lambda \begin{pmatrix}

x\\ 

y\end{pmatrix} = \begin{pmatrix}

1 &4 \\ 

2& 3

\end{pmatrix}\begin{pmatrix}

x\\ 

y

\end{pmatrix}

 \left (\left \begin{pmatrix} 1 &4 \\  2& 3 \end{pmatrix}-\begin{pmatrix} \lambda & 0 \\  0 & \lambda \end{pmatrix}  \right )\begin{pmatrix} x\\  y \end{pmatrix}=0

 det(A - \lambda I) \vec{v}=0
therefore
  ((1-\lambda )(3-\lambda )-8)=0
  \lambda =5  or \lambda = -1

I sub each  \lambda into the matrix;
  \begin{pmatrix} 1-\lambda & 4\\  2 & 3- \lambda \end{pmatrix}\begin{pmatrix} x\\  y \end{pmatrix} = 0

when  \lambda =5
 \begin{pmatrix} -4 & 4\\  2 & -2 \end{pmatrix}\begin{pmatrix} x\\  y \end{pmatrix} = 0

 -4x+4y=0
 2x-2y=0
thus y=x

when  \lambda = -1
 \begin{pmatrix} 2 & 4\\  2 & 4 \end{pmatrix}\begin{pmatrix} x\\  y \end{pmatrix} = 0
 2x +4y =0
  2x+ 4y = 0
thus   y = -\frac{1}{2}x


Did i follow the right approach?
2
reply
begbie68
Badges: 16
Rep:
?
#8
Report 3 weeks ago
#8
And the result in post#1 shows one of these:

the third column is transformed from column(1,1) to column(5,5) ... ie a point on the line y=x, (1,1) gets transformed to another point on the line y=x, (5,5) ...
1
reply
bigmansouf
Badges: 20
Rep:
?
#9
Report Thread starter 3 weeks ago
#9
yes i do thank you
(Original post by Ryanzmw)
Yep. Do you see how this relates to the question?
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

University open days

  • Solent University
    Postgraduate and Professional Open Evenings Postgraduate
    Mon, 20 May '19
  • London Metropolitan University
    Postgraduate Mini Open Evening - Holloway Campus Undergraduate
    Tue, 21 May '19
  • Brunel University London
    Postgraduate Open Evening Postgraduate
    Wed, 22 May '19

How has 2019 been so far?

Amazing!!! (39)
5.72%
Fairly positive (227)
33.28%
Just another year... (270)
39.59%
Is it 2020 yet? (146)
21.41%

Watched Threads

View All